Hey there!:
Molar mass:
CHCl3 = ( 12.01 * 1 )+ (1.008 * 1 ) + ( 35.45 * 3 ) => 119.37 g/mol
C% = ( atomic mass C / molar mass CHCl3 ) * 100
For C :
C % = (12.01 / 119.37 ) * 100
C% = ( 0.1006 * 100 )
C% = 10.06 %
For H :
H% = ( atomic mass H / molar mass CHCl3 ) * 100
H% = ( 1.008 / 119.37 ) * 100
H% = 0.008444 * 100
H% = 0.8444 %
For Cl :
Cl % ( molar mass Cl3 / molar mass CHCl3 ):
Cl% = ( 3 * 35.45 / 119.37 ) * 100
Cl% = ( 106.35 / 119.37 ) * 100
Cl% = 0.8909 * 100
Cl% = 89.9%
Hope that helps!
no the best source is blood.
Answer:
Find It Myself, Ask the Community, Get Live Help
Explanation:
The three main options for Microsoft users are Find It Myself, Ask the Community, Get Live Help. The first of which is finding it yourself through the search menu or help guidelines provided by Microsoft in all of their operating systems. Secondly, would be asking the community through search engines such as Google or through Microsoft help forums. Lastly, would be getting live help since Microsoft opertaing systems have a remote desktop feature that allows you to connect to another IT professional from a distance, or you can simply contact an IT professional to visit you in person for technical assistance.
Answer:
Covering electrical wires in plastic guarantees that the electrons flowing through the wires will not flow through your body when the wire is touched.
Explanation:
"The purpose of insulation covering the metal part of an electrical wire is to prevent accidental contact with other conductors of electricity, which might result in an unintentional electric current through those other conductors."
https://www.allaboutcircuits.com/worksheets/wire-types-and-sizes/#:~:text=Most%20electrical%20wire%20is%20covered,or%20plastic%20coating%20called%20insulation.&text=The%20purpose%20of%20insulation%20covering,current%20through%20those%20other%20conductors.
https://www.hunker.com/13414152/why-are-electrical-wires-covered-in-plastic
The reaction described above is the formation of an acetal. The initial starting material has a central carbonyl and two terminal alcohol functional groups. In the presence of acid, the carbonyl will become protonated, making the carbon of the carbonyl susceptible to nucleophilic attack from one of the alcohols. The alcohol substitutes onto the carbon of the carbonyl to provide us with the intermediate shown.
The intermediate will continue to react in the presence of acid and the -OH that was once the carbonyl will become protonated, turning it into a good leaving group. The protonated alcohol leaves and is substituted by the other terminal alcohol to give the final acetal product. The end result of the overall reaction is the loss of water from the original molecule to give the spiroacetal shown in the image provided.
